重现赛题目网站：(2条未读私信) 牛客竞赛_ACM/NOI/CSP/CCPC/ICPC算法编程高难度练习赛_牛客竞赛OJ
1.吃糖果(B组、C组)
#include<bits/stdc++.h>
#define int long long
using namespace std;
int n,k,count1=0,sum1=0;
int a[200010];
signed main()
{
cin>>n>>k;
for(int i=0;i<n;i++)
{
cin>>a[i];
}
sort(a,a+n);
for(int i=0;i<n;i++)
{
sum1+=a[i];
if(sum1<=k)
{
count1++;
}
else
{
break;
}
}
cout<<count1<<endl;
return 0;
}2.汤姆和杰瑞(A组、C组)
#include<iostream>
#define int long long
using namespace std;
signed main()
{
int a,b;
cin>>a>>b;
cout<<b-a<<" "<<b<<endl;
return 0;
}3.游游的重组偶数(A组、B组、C组)
发现偶数规律，如果用整形重组注意前导0的情况
#include<bits/stdc++.h>
#define int long long
using namespace std;
int q;
signed main()
{
cin>>q;
while(q--)
{
int x;
cin>>x; //变为偶数，数位存在偶数就行
int a=0,x1=0,temp=0;;
bool found=false;
string s,s1;
while(x>0)
{
a=x%10;
x/=10;
if(a%2==0)
{
found=true;
temp=a;
break;
}
x1=x1*10+a;
s=to_string(x1);
}
if(x!=0)
{
s1=s+to_string(x)+to_string(temp);
}
else
{
s1=s+to_string(temp);
}
int temp1=0;
for(int i=0;i<s1.size();i++)
{
temp1=temp1*10+(int)(s1[i]-'0');
}
if(found)
{
cout<<temp1<<endl;
}
else
{
cout<<-1<<endl;
}
}
return 0;
}4.开心还是难过(B组、C组)
注意处理空格字符串的方式：getline(cin,string);
#include<bits/stdc++.h>
#define int long long
using namespace std;
string str;
signed main()
{
getline(cin,str);
int s1=0,s2=0;
for(int i=0;i<str.size();i++)
{
if(str[i]==':'&&str[i+1]=='-'&&str[i+2]=='(')
{
s1++;
}
else if((str[i]==':'&&str[i+1]=='-'&&str[i+2]==')'))
{
s2++;
}
}
if((s1==0)&&(s2==0))
{
cout<<"None"<<endl;
}
else if(s1<s2)
{
cout<<"Happy"<<endl;
}
else if(s1==s2)
{
cout<<"Just so so"<<endl;
}
else if(s1>s2)
{
cout<<"Sad"<<endl;
}
return 0;
}5.小欧的平面连线(A组、B组、C组)
规律题，交叉对角象限权值2，交叉单坐标权值1
#include<iostream>
#define int long long
using namespace std;
int n,sum=0;
signed main()
{
int x,y,sign1=0,sign2=0,sign3=0,sign4=0;
cin>>n;
while(n--)
{
cin>>x>>y;
if(x>=0 && y>=0)
{
sign1++; //一象限
}
else if(x<=0 && y>=0)
{
sign2++;
}
else if(x<=0 && y<=0)
{
sign3++;
}
else
{
sign4++;
}
}
int temp,temps,temp1,temp2;
//1,3象限配对 2,4象限配对
temp=min(sign1,sign3);
temp1=max(sign1,sign3)-temp; //1,3象限剩余某单个象限点
temps=min(sign2,sign4);
temp2=max(sign2,sign4)-temps;
sum=(temp+temps)*2;
sum+=min(temp1,temp2); //权值为1配对
cout<<sum<<endl;
return 0;
}6.小红的四子棋(A组、B组、C组)
遍历数组，注意if (a[i][j] == '.') continue; // 忽略
只需要判断有4个棋子相连即可，更多无意义
#include<bits/stdc++.h>
#define int long long
using namespace std;
int n,m;
char a[110][110];
signed main()
{
cin>>n>>m;
for(int i=1; i<=n; i++)
{
for(int j=1; j<=m; j++)
{
cin>>a[i][j];
}
}
char a1='1';
for(int i=1; i<=n; i++)
{
for(int j=1; j<=m; j++)
{
if (a[i][j] == '.') continue; // 忽略
if(a[i][j]==a[i+1][j] && a[i][j]==a[i+2][j] && a[i][j]==a[i+3][j])
{
if(a[i][j]=='r')
{
cout<<"kou"<<endl;
}
else
{
cout<<"yukari"<<endl;
}
return 0;
}
else if(a[i][j]==a[i][j+1] && a[i][j]==a[i][j+2] && a[i][j]==a[i][j+3])
{
if(a[i][j]=='r')
{
cout<<"kou"<<endl;
}
else
{
cout<<"yukari"<<endl;
}
return 0;
}
else if(a[i][j]==a[i+1][j+1] && a[i][j]==a[i+2][j+2] && a[i][j]==a[i+3][j+3])
{
if(a[i][j]=='r')
{
cout<<"kou"<<endl;
}
else
{
cout<<"yukari"<<endl;
}
return 0;
}
else if(a[i][j]==a[i+1][j-1] && a[i][j]==a[i+2][j-2] && a[i][j]==a[i+3][j-3] )
{
if(a[i][j]=='r')
{
cout<<"kou"<<endl;
}
else
{
cout<<"yukari"<<endl;
}
return 0;
}
}
}
cout<<"to be continued"<<endl;
return 0;
}7.小红的数组操作(A组、B组)
二分思想，二分寻找 mid:数组尽可能小的最大值 ，注意可以是负数
#include<iostream>
#define int long long
using namespace std;
int a[100010];
int n,k,x;
bool check(int mid)
{
int sum=0; //操作次数
for(int i=1; i<=n; i++)
{
if(a[i]>mid) //计算将a[i]降到mid需要的操作次数
{
sum+=(a[i]-mid+x-1)/x; //总共操作次数
}
}
return sum<=k; //非必等于k次，尽可能小的最大值（小于k次就能实现，剩余操作只要不去操作数组最大值就行）
}
signed main()
{
cin>>n>>k>>x;
for(int i=1; i<=n; i++) cin>>a[i];
int l = -1e14, r = 1e9, mid, ans;
while(l<=r)
{
mid=(l+r)/2;
if(check(mid)) //mid:数组尽可能小的最大值
{
ans=mid;
r=mid-1; //寻找更小的mid
}
else
{
l=mid+1;
}
}
cout<<ans<<endl;
return 0;
}